Question

A buffer is made by adding 0.350 mol lactic acid, CH₃CH(OH)COOH, and 0.350 mol potassium lactate, CH₃CH(OH)COOK, to enough water to make 1.10 L of solution. The pKa is equal to 4.24. Calculate the pH of this solution after 3.0 mL of 3.0 M KOH is added to the buffer.

Answer 1

So, Moles of CH3CH(OH)CO2H Moles of CH3CH(OH)CO2- = = 0.3500 0.3500 mol mol pH after OH- addition to the buffer Moles of Base added = Molarity (M) * Volume (V in L) So, Moles of [OH-] added = 3.000 M * 0.0030 L = 0.0090 mol When OH- is added to the buffer, it reacts with acid CH3CH(OH)CO2H and produce CH3CH(OH)CO2-. So during the addition of OH-to the buffer, CH3CH(OH)CO2- Conc/moles will increase and CH3CH(OH)CO2H moles / conc will decrease as shown in the below table. <-------> Net ionic equation after addition of OH-to the Buffer CH3CH(OH)CO2H (aq) + OH- (aq) Initial 0.350 0.0090 Change -0.0090 - 0.0090 Equil 0.341 0.000 Henderson equation for buffer solution is CH3CH(OH)CO2- (aq) 0.350 + 0.0090 0.359 log [ Base Acid Acid ] pH = pka Here the Base is CH3CH(OH)CO2- and Acid is CH3CH(OH)CO2H pH = 4.240 + log 1- log 0.3590 0.34101 0.3410 1.0528 + 4.240 4.240 4.26 log! 0.0223 = + 4.262 pH of buffer after addition of (OH-)